1. It is stated "A 10 Ohm load resistor is soldered in positions P1, P2, P3, P4 and P5 on the terminal board. " I have calculated the Rload I need, in the terminal board came along with ADC20, the Rload should be soldered in positions R1b? Am I right?

2. It is stated "Batteries, in holders, are then connected to channels 1 to 5." For the terminal board came along with ADC20, where should I connect the Lithium Ion cell (3.75V, 5300mAh) to?

3. I want to measure Voltage and Current both at the same time. In the ADC-20/ADC-24 Terminal Board User Guide, page 4 to 6, as the voltage I am going to use is 4.2 V(Which goes beyond 2.5 V), so I am going to use voltage divider connection. I have calculated the RA (5.95 k ohm) and RB (4.05 k ohm) values for voltage measurement as well as RB2 (0.5 ohm) value for Current measurement. My question is, how am I going to put these three resistors on the Terminal Board? Is it true that RA positioned at R1a, RB at R1b and RB2 at R2b?

4. Other than discharging the battery, I need to charge the battery as well, so I am going to connect Power Supply to Terminal Board came along with ADC-20, where should I position the Power Supply?

Hi,
The terminal board for the ADC-20 is completely different to that for the ADC-16 and I would not advise trying to add the various load resistors needed for your testing to this terminal board. The batteries and load resistors plus any switches and charging connections should be put together on a separate assembly. The voltages to be measured by the ADC-20 should be connected to the terminal board by simple wire connections between your battery assembly and the terminals of the ADC-20 terminal board.
I suggest that you use two load resistors in series across each battery to both supply the load and also attenuate the voltage to the correct level to feed to the ADC-20. If you want 10 ohms load for each battery, use two 4.7ohm resistors in series. That will load the batteries with 9.7 ohms and the voltage across the resistor connected to the -Ve terminal of the battery will be half battery voltage at 1.85V. Multiply that voltage by 2 using the scaling function of PicoLog to get the battery voltage. Take that same 1.85V and divide it by 4.7 ohms and that will give you the current drawn by the load resistors.
Please contact me again if you have any other questions.
Regards,
PeterF.

1. How could I measure voltage and current at the SAME time?
Because the RB values are different. For example, RB (4.05 k ohm) values for voltage measurement and RB (0.5 ohm) value for Current measurement.

2. I didn't get the reason behind this: " If you want 10 ohms load for each battery, use two 4.7ohm resistors in series. That will load the batteries with 9.7 ohms and the voltage across the resistor connected to the -Ve terminal of the battery will be half battery voltage at 1.85V. " Mind to explain?

3. What is the connecting points between battery assembly and the terminals of the ADC-20 terminal board?

4. If I follow the methodology you proposed, for voltage measurement, do I still need to have the voltage divider connections on the ADC-20 terminal board as stated in the instruction manual?

5. If I follow the methodology you proposed, for current measurement, do I still need to have the shunt resistor to convert voltage into current on the ADC-20 terminal board as stated in the instruction manual?

Hi,
What I proposed is completely different to the original application note. Using two 4.7 ohm resistors in series connected across each battery allows measurement of voltage and current at the same time. By measuring the voltage across the resistor connected to 0V, the measured voltage will be half the battery voltage since the two resistors form a 2:1 attenuator. If this voltage is multiplied by two in PicoLog the battery voltage will be shown. The two 4.7 ohm resistors also act as loads for the battery, drawing current. From ohms law, V/R=I, so 3.75V divided by 9.4 ohms equals 399mA, nominally. To calculate the actual current at any point, take the voltage across the one resistor, divide by the value of the one resistor and that will give you the current drawn from the battery. To plot both voltage and current at the same time in PicoLog, connect two logger channels to each monitored resistor with one channel set to display full battery voltage and the other to display battery current. I can help you with the set-up of PicoLog to get the required scaling, if you wish.
No other resistors or attenuators are required. Connection is direct to the logger input channels.
Please contact me again if you have any other questions.
Regards,
PeterF.

Thank you very much for the detailed explanation, Sir.
1. I now understand the reason behind this. If I understand correctly, this can only be applied for discharging. For voltage and current measurement during Charging purpose, I don't connect to load, but to power supply instead, so how should I do the measurement?
2. Yes. Please help me with the set up of Picolog for required scaling.
For current measurement, since we measure the voltage then only convert to current, so I have no idea how to do the set up at Picolog scaling.
Thank you in advance

I am going to measure THREE parameters at the SAME time: voltage, current and temperature (using LM35DZ IC).

In the Terminal Board, there are 16 channels totally (from Channel 1 to Channel 16), I have no problem with this since I only need three and there are 16 available.

My question is, How am I going to do the setting to use THREE channels in PICOLOG?

In File> New Settings > Measurement Tab > Add
I repeated the above steps for three times, but in the "Edit measurement" Tab, Under "Channel", only two channels are available to be selected from the list (Channel A and Channel B). I dont know what these Channel A and Channel B mean.

In File> New Settings > Recording, should I check or uncheck "Use multiple converters" in my application?

Hi,
The fact that you are only seeing two channels, A & B, suggests to me that when you installed PicoLog on your PC that you have selected as your device "PicoScope 2000 series" and you did not tick the box for "Install for other devices". Please re-run your PicoLog installation file, select "Remove" when you are asked to select "Repair, Modify or Remove". After removal, re-run the installer to install a fresh copy and when asked "For which device do you wish to install" select "ADC-20/24". Also, for maximum compatibility, tick the small box for "Install for other devices".
You should then be able to run PicoLog and see all 8 channels of the ADC-20.
If you wish to monitor current and voltage during both charge and discharge you will need a different circuit layout again. Please fit two 10k resistors to the terminal board for each channel that to wish to use for voltage measurement. As shown on the circuit sketch t hat I attach, R1A & R1B for channel 1. (Do not forget to cut the track underneath R1A.) This will halve the voltage seen by the ADC-20 channel and allow +5V measurements. For the rest of the circuit, you will require a 9.1 ohm discharge load resistor, a 1 ohm current sense resistor and a switch to change over from charge to discharge testing. Wire everything up as per the diagram. In use, you should see half the actual battery voltage on channel 1 and a voltage on channel two such that 1 V equals 1A. If you get that, all is working well. Please contact me then and I will tell you how to set the PicoLog scaling to suit. The current channel will show a +Ve value for discharge and a -Ve value for charge (This can be inverted in the scaling if you wish. If the current for either charge or discharge exceeds 2.5A, then the value of the current sense 1 ohm resistor will need to be lower.
Please contact me again if you have any other questions.
Regards,
PeterF.

Hi,
If you have separate charge/discharge changeover switches for each battery then you can have one set on charge and one set on discharge at the same time. It does not matter to the ADC-20, it just measures the current, in or out of the batteries.
All the AG terminals of the terminal board are connected together as one common point so it does not matter which you connect to. Negative terminals of BOTH Battery A and Battery B are at the same potential.
Please contact me again if you have any other questions.
Regards,
PeterF.